Question

a proton is accelerating in a cyclotron where the applied magnetic field is 2T. if the potential gap is effectively 100kV then how much revolutions the proton has to make between the "dees" to acquire energy of 20MeV?​

Answer 1

Answer:

Given, Magnetic field B=2T

Given, Magnetic field B=2TPotential gap ΔV=100kV

Given, Magnetic field B=2TPotential gap ΔV=100kVWe know that, r=qBmv

Given, Magnetic field B=2TPotential gap ΔV=100kVWe know that, r=qBmv⇒V=100×10^3=10^5 volts

 voltsSo, KE gained in each revolution  =e(V)+e(V)=2e(V)=2e×10^5

Thus, number of revolution    N=(20×10^6)/(2×10^5) =100revolution

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